
(a)
Interpretation:Numbers of possible stereoisomeric products formed in the monobromination of racemic
Concept introduction:Halogenation products are governed by the achirality or already present stereocenter in the molecule.
If the halogenation occurs at the chiral center of the substrate it results in a racemic mixture. This is attributed to formation of achiral radical intermediate that can be attacked equally from the top face or bottom face. Thus two enantiomers R and S are formed in equal amount that makes the overall mixture optically inactive.
If the halogenation occurs at one of the two chiral centers of the substrate it results in change in absolute configuration at only one of the stereocenter and thus diastereoisomer result. These diastereomers are formed in unequal amounts.
(b)
Interpretation: Numbers of possible stereoisomeric products formed in the monobromination of racemic
Concept introduction:Halogenation products are governed by the achirality or already present stereocenter in the molecule.
If the halogenation occurs at the chiral center of the substrate it results in a racemic mixture. This is attributed to formation of achiral radical intermediate that can be attacked equally from the top face or bottom face. Thus two enantiomers R and S are formed in equal amount that makes the overall mixture optically inactive.
If the halogenation occurs at one of the two chiral centers of the substrate it results in change in absolute configuration at only one of the stereocenter and thus diastereoisomer result. These diastereomers are formed in unequal amounts.
(c)
Interpretation:Whether the various stereoisomeric products obtained from monobromination of racemic
Concept introduction:Halogenation products are governed by the achirality or already present stereocenter in the molecule.
If the halogenation occurs at the chiral center of the substrate it results in a racemic mixture. This is attributed to formation of achiral radical intermediate that can be attacked equally from the top face or bottom face. Thus two enantiomers R and S are formed in equal amount that makes the overall mixture optically inactive.
If the halogenation occurs at one of the two chiral centers of the substrate it results in change in absolute configuration at only one of the stereocenter and thus diastereoisomer result. These diastereomers are formed in unequal amounts.

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Chapter 5 Solutions
ORGANIC CHEMISTRY (LL)-PACKAGE
- Steps and explanations pleasearrow_forwardUse diagram to answer the following: 1.Is the overall rxn endo- or exothermic. Explain briefly your answer____________________2. How many steps in this mechanism?_____________3. Which is the rate determining step? Explain briefly your answer____________________4. Identify (circle and label) the reactants,the products and intermediate (Is a Cation, Anion, or a Radical?) Please explain and provide full understanding.arrow_forwardDraw the entire mechanism and add Curved Arrows to show clearly how electrons areredistributed in the process. Please explain and provide steps clearly.arrow_forward
- Match the denticity to the ligand. Water monodentate ✓ C₂O2 bidentate H₂NCH₂NHCH2NH2 bidentate x EDTA hexadentate Question 12 Partially correct Mark 2 out of 2 Flag question Provide the required information for the coordination compound shown below: Na NC-Ag-CN] Number of ligands: 20 Coordination number: 2✔ Geometry: linear Oxidation state of transition metal ion: +3 x in 12 correct out of 2 question Provide the required information for the coordination compound shown below. Na NC-Ag-CN] Number of ligands: 20 Coordination number: 2 Geometry: linear 0 Oxidation state of transition metal ion: +3Xarrow_forwardCan you explain step by step behind what the synthetic strategy would be?arrow_forwardPlease explain step by step in detail the reasoning behind this problem/approach/and answer. thank you!arrow_forward
- 2. Predict the product(s) that forms and explain why it forms. Assume that any necessary catalytic acid is present. .OH HO H₂N OHarrow_forwardconsider the rate of the reaction below to be r. Whats the rate after each reaction? Br + NaCN CN + NaBr a. Double the concentration of alkyl bromide b. Halve the concentration of the electrophile & triple concentration of cyanide c. Halve the concentration of alkyl chloridearrow_forwardPredict the organic reactant that is involved in the reaction below, and draw the skeletal ("line") structures of the missing organic reactant. Please include all steps & drawings & explanations.arrow_forward
- Macroscale and Microscale Organic ExperimentsChemistryISBN:9781305577190Author:Kenneth L. Williamson, Katherine M. MastersPublisher:Brooks Cole

